2002 AMC 10A Problems/Problem 16

Problem

Let $\text{a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}$ . What is $\text{a + b + c + d}$ ?

$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$

Solution

Let $\text{x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}$ . Since one of the sums involves a,b,c, and d, it makes sense to consider 4x. $\text{4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20}$ . Rearranging, we have $\text{3(a+b+c+d)=-10}$ , so $\text{a+b+c+d}=\frac{-10}{3}$ . Thus, our answer is $\boxed{\text{(B)}\ -10/3}$ .

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2002 AMC 10A Problems/Problem 16

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